Infimum of Set of Reciprocals of Positive Integers

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Theorem

Let $S$ be the subset of the set of real numbers defined as:

$S = \set {\dfrac 1 n: n \in \Z_{>0} }$

Then:

$\inf S = 0$

where $\inf S$ denotes the infimum of $S$.


Proof

We have that $\Z_{>0}$ contains only (strictly) positive integers.

So it follows from Reciprocal of Positive Real Number is Positive that $S$ contains only (strictly) positive real numbers.

Hence $0$ is a lower bound for $S$.


Aiming for a contradiction, suppose that $0$ is not the infimum of $S$.

Then $\exists h \in \R_{>0}$ such that $h$ is a lower bound for $S$.

Then:

$\forall n \in \Z_{>0}: \dfrac 1 n \ge h$

Hence from Reciprocal Function is Strictly Decreasing:

$\forall n \in \Z_{>0}: n \le \dfrac 1 h$

where $\dfrac 1 h \in \R$.

But from the Archimedean Principle:

$\exists n \in \Z_{>0}: n > \dfrac 1 h$

This is a contradiction.

It follows by Proof by Contradiction that no such $h > 0$ exists.

Hence $0$ is the infimum of $S$.

$\blacksquare$


Sources